# 6.4.1: Circles Centered at the Origin

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# Circles Centered at the Origin

You draw a **circle** that is centered at the origin. You measure the **diameter** of the circle to be 32 units. Does the point \(\ (14,8)\) lie on the circle?

# Circles Centered at the Origin

Until now, your only reference to circles was from geometry. A ** circle** is the set of points that are equidistant (the

**) from a given point (the**

**radius****). A line segment that passes through the center and has endpoints on the circle is a**

**center****.**

**diameter**Now, we will take a circle and place it on the x−y plane to see if we can find its equation. In this concept, we are going to place the center of the circle on the origin.

# Finding the Equation of a Circle

Step 1: On a piece of graph paper, draw an x−y plane. Using a compass, draw a circle, centered at the origin that has a radius of 5. Find the point (3,4) on the circle and draw a right triangle with the radius as the hypotenuse.

Step 2: Using the length of each side of the right triangle, show that the Pythagorean Theorem is true.

Step 3: Now, instead of using (3, 4), change the point to (x, y) so that it represents any point on the circle. Using r to represent the radius, rewrite the Pythagorean Theorem.

The ** equation of a circle**, centered at the origin, is \(\ x^{2}+y^{2}=r^{2}\), where r is the radius and (x, y) is any point on the circle.

Let's find the radius of \(\ x^{2}+y^{2}=16\) and graph.

To find the radius, we can set \(\ 16=r^{2}\), making \(\ r=4\). \(\ r\) is not -4 because it is a distance and distances are always positive. To graph the circle, start at the origin and go out 4 units in each direction and connect.

Now, let's find the equation of the circle with center at the origin and passes through \(\ (−7,−7)\).

Using the equation of the circle, we have: \(\ (-7)^{2}+(-7)^{2}=r^{2}\). Solve for \(\ r^{2}\).

\(\ \begin{aligned}

(-7)^{2}+(-7)^{2} &=r^{2} \\

49+49 &=r^{2} \\

98 &=r^{2}

\end{aligned}\)

So, the equation is \(\ x^{2}+y^{2}=98\). The radius of the circle is \(\ r=\sqrt{98}=7 \sqrt{2}\).

Finally, let's determine if the point \(\ (9,−11)\) is on the circle \(\ x^{2}+y^{2}=225\).

Substitute the point in for x and y and see if it equals 225.

\(\ \begin{aligned}

9^{2}+(-11)^{2} &=225 \\

81+121 & \stackrel{?}{=} 225 \\

202 & \neq 225

\end{aligned}\)

The point is not on the circle.

# Examples

Example 1

Earlier, you were asked to determine if the point (14 ,8) lies on the circle that is centered at the origin and has a diameter of 32 units.

**Solution**

From this lesson, you know that the equation of a circle that is centered at the origin is \(\ x^{2}+y^{2}=r^{2}\), where \(\ r\) is the radius and \(\ (x,y)\) is any point on the circle.

With the point \(\ (14,8)\), \(\ x=14\) and \(\ y=8\). We are given the diameter, but we need the radius. Recall that the radius is half the diameter, so the radius is \(\ \frac{32}{2}=16\).

Plug these values into the equation of the circle. If they result in a true statement, the point lies on the circle.

\(\ \begin{array}{r}

x^{2}+y^{2}=r^{2} \\

14^{2}+8^{2} \stackrel{?}{=} 16^{2} \\

196+64 \stackrel{?}{=} 256 \\

260 \neq 256

\end{array}\)

Therefore the point does not lie on the circle.

Example 2

Graph and find the radius of \(\ x^{2}+y^{2}=4\).

**Solution**

\(\ r=\sqrt{4}=2\)

Example 3

Find the equation of the circle with a radius of \(\ 6 \sqrt{5}\).

**Solution**

Plug in \(\ 6 \sqrt{5}\) for \(\ r\) in \(\ x^{2}+y^{2}=r^{2}\)

\(\ \begin{array}{l}

x^{2}+y^{2}=(6 \sqrt{5})^{2} \\

x^{2}+y^{2}=6^{2} \cdot(\sqrt{5})^{2} \\

x^{2}+y^{2}=36 \cdot 5 \\

x^{2}+y^{2}=180

\end{array}\)

Example 4

Find the equation of the circle that passes through (5, 8).

**Solution**

Plug in (5, 8) for x and y, respectively.

\(\ \begin{aligned}

5^{2}+8^{2} &=r^{2} \\

25+64 &=r^{2} \\

89 &=r^{2}

\end{aligned}\)

The equation is \(\ x^{2}+y^{2}=89\)

Example 5

Determine if \(\ (−10, 7)\) is on the circle \(\ x^{2}+y^{2}=149\).

**Solution**

Plug in \(\ (−10, 7)\) to see if it is a valid equation.

\(\ \begin{aligned}

(-10)^{2}+7^{2} &=149 \\

100+49 &=149

\end{aligned}\)

Yes, the point is on the circle.

# Review

Graph the following circles and find the radius.

- \(\ x^{2}+y^{2}=9\)
- \(\ x^{2}+y^{2}=64\)
- \(\ x^{2}+y^{2}=8\)
- \(\ x^{2}+y^{2}=50\)
- \(\ 2 x^{2}+2 y^{2}=162\)
- \(\ 5 x^{2}+5 y^{2}=150\)

Write the equation of the circle with the given radius and centered at the origin.

- \(\ 14\)
- \(\ 6\)
- \(\ 9 \sqrt{2}\)

Write the equation of the circle that passes through the given point and is centered at the origin.

- \(\ (7,−24)\)
- \(\ (2,2)\)
- \(\ (−9,−10)\)

Determine if the following points are on the circle, \(\ x^{2}+y^{2}=74\).

- \(\ (−8,0)\)
- \(\ (7,−5)\)
- \(\ (6,−6)\)

** Challenge** In Geometry, you learned about tangent lines to a circle. Recall that the tangent line touches a circle at one point and is perpendicular to the radius at that point, called the point of tangency.

- The equation of a circle is \(\ x^{2}+y^{2}=10\) with point of tangency \(\ (−3,1)\).
- Repeat the steps in #16 to find the equation of the tangent line to \(\ x^{2}+y^{2}=34\) with a point of tangency of \(\ (3,5)\).

# Answers for Review Problems

To see the Review answers, open this PDF file and look for section 10.3.

# Vocabulary

Term | Definition |
---|---|

center |
The center of a circle is the point that defines the location of the circle. All points on the circle are equidistant from the center of the circle. |

Circle |
A circle is the set of all points at a specific distance from a given point in two dimensions. |

Diameter |
Diameter is the measure of the distance across the center of a circle. The diameter is equal to twice the measure of the radius. |

Equation of a Circle |
If the center of a circle is (0, 0), then the equation of the circle is of the form \(\ x^{2}+y^{2}=r^{2}\), where \(\ r\) is the radius. |

Radius |
The radius of a circle is the distance from the center of the circle to the edge of the circle. |

# Image Attributions

- [Figure 1]

Tony Webster;Robert Pernett**Credit:**

https://commons.wikimedia.org/wiki/File%3ALas_Vegas_High_Roller_(20216869960).jpg;https://flic.kr/p/nQD1fo**Source:**